I'm dealing with the following problem.
Suppose we have a random variable $X\in R^n$ and a function $g(X)$ with $g(x)$ for $X = x$. We assume $g(x)$ is continuous in x and the marginal density $f(x)$ is also continuous in x.
Also define $B_\epsilon(x)$, the ball around $x$ of radius $\epsilon$ according to the euclidean distance.
We are interested in a continuous curve $B$, for example a line, so we also define a "neighborhood" around it as $B_\epsilon = \cup_{x_0 \in B}B_\epsilon(x_0)$.
We want to prove:
$\lim_{\epsilon \to 0 }\frac{\int_{B_\epsilon(x_0)} g(x)f(x)dx}{ \int_{B_\epsilon(x_0)}f(x)dx} = g(x_0)$
$\lim_{\epsilon \to 0}\frac{\int_{B_\epsilon} g(x)f(x)dx}{ \int_{B_\epsilon}f(x)dx} = \frac{\int_{B} g(x)f(x)dx}{ \int_{B}f(x)dx}$
I think I managed to prove the first one in this way:
Because $g(x)$ is continuous for all $x$ then for all $\delta$ $\exists \epsilon$ such for all $x \in B_\epsilon(x_0)$ we have
$| g(x) - g(x_0)| < \delta$
and
$\frac{\int_{B_\epsilon(x_0)} |g(x) - g(x_0)|f(x)dx}{\int_{B_\epsilon(x_0)} f(x)dx} < \frac{\int_{B_\epsilon(x_0)} \delta f(x)dx}{\int_{B_\epsilon(x_0)} f(x)dx} = \delta$.
However, I can't figure out the second statement. I tried many routes to no avail. Can anyone help me or point to the right references? Thank you!
Let $$\varphi_\varepsilon := \chi_{B_\varepsilon} f g.$$ Clearly, $\varphi_\varepsilon$ is measurable, $\vert \varphi_\varepsilon \vert$ is decreasing in $\varepsilon$ and $\lim_{\varepsilon \to 0} \varphi_\varepsilon = \chi_B f g.$ Let us discretize this sequence by denoting $\varphi_n := \varphi_{\frac{1}{n}}$ and $B_n := B_{\frac{1}{n}}.$ Then $$\int_{B_n} f g d \lambda = \int \chi_{B_n} f g d \lambda = \int \varphi_n d \lambda.$$ We can see that $\vert \chi_{B_n} f g \vert \leq \vert f g \vert,$ so the Dominated Convergence Theorem yields that $\int \varphi_n d \lambda \to \int_B f g d \lambda$ (as long as $\int_{B_n} \vert f g \vert d \lambda < \infty$ for some $n \in \mathbb{N}$). Mutatis mutandis, $\int_{B_n} f d \lambda = \int \chi_{B_n} f d \lambda \to \int_B f d \lambda.$ Putting the two results together concludes our proof. I hope this helps. :)